In the next questions, a particle is moving along a horizontal line according to the formula:
s=2t^4-4t^3+2t^2-1
a) the particle is moving right when
A. 0 is less than t is less than 1/2
B. t is greater than 0
C. t is greater than 1
D. 0 is less than t is less than 1/2, t is greater than 1
E. Never
b) the acceleration, a is increasing when
A. t is greater than 1
B. t is greater than 0.5
C. t is less than 0.211 or t is greater than .789
D. 0 is less than t is less than 0.5
E. 0 is less than t is less than 1
moving right when s is increasing. That is, when ds/dt > 0
a is increasing when da/dt is positive. That is,
48t-24 > 0
t > 2
t is greater than 0.5, not 2
a) To determine when the particle is moving right, we need to find the values of t for which the velocity of the particle is positive. The velocity, v(t), is the derivative of the position function, s(t):
v(t) = 8t^3 - 12t^2 + 4t
Now we can find the critical points of v(t) by setting it equal to zero and solving for t:
8t^3 - 12t^2 + 4t = 0
Factor out a common factor of 4t:
4t(2t^2 - 3t + 1) = 0
Now we have two possibilities:
1. 4t = 0, which gives t = 0.
2. 2t^2 - 3t + 1 = 0, which can be factored as (2t - 1)(t - 1) = 0. This gives t = 1/2 or t = 1.
Now we need to determine the sign of v(t) for different values of t:
For t < 0, v(t) = 8t^3 - 12t^2 + 4t < 0 (since all the terms are negative).
For 0 < t < 1/2, v(t) = 8t^3 - 12t^2 + 4t > 0 (since all the terms are positive).
For 1/2 < t < 1, v(t) = 8t^3 - 12t^2 + 4t < 0 (since the first term is positive and the other terms are negative).
For t > 1, v(t) = 8t^3 - 12t^2 + 4t > 0 (since all the terms are positive).
Therefore, the particle is moving right when 0 < t < 1/2 and t > 1. The correct answer is D. 0 is less than t is less than 1/2, t is greater than 1.
b) To determine when the acceleration is increasing, we need to find the values of t for which the acceleration is positive and its derivative (change in acceleration) is also positive.
The acceleration, a(t), is the derivative of the velocity function, v(t):
a(t) = 24t^2 - 24t + 4
Now we can find the critical points of a(t) by setting its derivative equal to zero and solving for t:
a'(t) = 48t - 24 = 0
Simplifying, we find t = 0.5.
Next, we need to determine the sign of a(t) and a'(t) for different values of t:
For t < 0.5, a(t) = 24t^2 - 24t + 4 > 0 (since all the terms are positive) and a'(t) = 48t - 24 > 0 (since t < 0.5).
For t > 0.5, a(t) = 24t^2 - 24t + 4 > 0 (since all the terms are positive) and a'(t) = 48t - 24 > 0 (since t > 0.5).
Therefore, the acceleration, a, is increasing when t is greater than 0.5. The correct answer is B. t is greater than 0.5.
To determine when the particle is moving right, we need to analyze the velocity equation, which is the derivative of the position equation.
The velocity equation v(t) can be found by taking the derivative of the position equation s(t) with respect to t:
v(t) = d/dt (2t^4 - 4t^3 + 2t^2 - 1)
To find when the particle is moving right, we need to determine when the velocity is positive (v(t) > 0). Let's solve for v(t) > 0:
2t^4 - 4t^3 + 2t^2 - 1 > 0
Now we can factor this equation as follows:
(t - 1)(2t + 1)(t^2 - t + 1) > 0
Let's solve each factor separately:
1. (t - 1) > 0
This implies t > 1.
2. (2t + 1) > 0
This implies t > -1/2.
3. (t^2 - t + 1) > 0
To determine the sign of this quadratic equation, we can either solve it explicitly or use the fact that the discriminant (b^2 - 4ac) is negative because the coefficient of t^2 is 1, the coefficient of t is -1, and the constant term is 1. Therefore, the quadratic equation is positive for all real values of t.
Now we need to find the intersection points of these three intervals: (-∞, -1/2), (-1/2, 1), and (1, +∞). We can pick a test point from each interval to see if it satisfies v(t) > 0:
1. Choose t = -1. Since -1 is in the interval (-∞, -1/2), substitute it into the equation:
(t - 1)(2t + 1)(t^2 - t + 1) = (-1 - 1)(2(-1) + 1)((-1)^2 - (-1) + 1) = (-2)(-1)(1) = 2 > 0.
2. Choose t = 0. Since 0 is in the interval (-1/2, 1), substitute it into the equation:
(t - 1)(2t + 1)(t^2 - t + 1) = (0 - 1)(2(0) + 1)(0^2 - 0 + 1) = (-1)(1)(1) = -1 < 0.
3. Choose t = 2. Since 2 is in the interval (1, +∞), substitute it into the equation:
(t - 1)(2t + 1)(t^2 - t + 1) = (2 - 1)(2(2) + 1)((2)^2 - 2 + 1) = (1)(5)(3) = 15 > 0.
Based on these test points, we can conclude that the particle is moving right when t is in the interval (-∞, -1/2) ∪ (1, +∞).
Therefore, the correct answer to part (a) is E. Never.
Now let's move on to part (b) and determine when the acceleration is increasing. The acceleration equation a(t) is the derivative of the velocity equation v(t).
a(t) = d/dt (v(t))
To find when the acceleration is increasing, we need to analyze the second derivative of the position equation.
Let's find a(t) by taking the derivative of v(t):
a(t) = d^2/dt^2 (2t^4 - 4t^3 + 2t^2 - 1)
Taking the second derivative, we have:
a(t) = 12t^2 - 24t + 4
To find when the acceleration is increasing, we need to determine when the second derivative is positive (a(t) > 0).
Setting the equation a(t) > 0, we have:
12t^2 - 24t + 4 > 0
To find when this quadratic equation is positive, we can either solve it explicitly or analyze the sign of the discriminant.
Analyzing the discriminant:
b^2 - 4ac = (-24)^2 - 4(12)(4) = 576 - 192 = 384 > 0
Since the discriminant is positive, the quadratic equation is positive for all real values of t.
Therefore, the acceleration is increasing for all values of t.
Based on this analysis, the correct answer to part (b) is E. Never.